সম্প্রসারণ: বাটারফ্লাই উপপাদ্য

The Butterfly Theorem

Draw chords AB and CD through the midpoint M of a chord PQ in a circle. Let AD and BC intersect chord PQ at points X and Y respectively. Then, M is the midpoint of XY.

Proof

(credit to cut the knot)

First, let $a=MP=MQ$ and $x=XM$ and $y=YM$. We drop perpendiculars from X and Y to AM, DM, and BM, CM, respectively. We call these perpendiculars $XX_1, XX_2, YY_1,$ and $YY_2,$ respectively, with lengths $x_1, x_2, y_1, y_2$.

We see four pairs of similar triangles, and we get a ratio from each: $$MXX_1 \sim MYY_1$$ $$\dfrac xy = \dfrac{x_1}{y_1}$$ $$MXX_2 \sim MYY_2$$ $$\dfrac xy = \dfrac{x_2}{y_2}$$ $$AXX_1 \sim CYY_2$$ $$\dfrac{x_1}{y_2} = \dfrac{AX}{CY}$$ $$DXX_2 \sim BYY_1$$ $$\dfrac{x_2}{y_1} = \dfrac{XD}{YB}$$ Then, from the above expressions, we get: $$\dfrac{x^2}{y^2} = \dfrac{x_1}{y_1} \cdot \dfrac{x_2}{y_2} = \dfrac{x_1}{y_2} \cdot \dfrac{x_2}{y_1} = \dfrac{AX \cdot XD}{CY \cdot YB}$$ By power of a point, $$\dfrac{AX \cdot XD}{CY \cdot YB} = \dfrac{PX \cdot XQ}{PY \cdot YQ} = \dfrac{(a-x)(a+x)}{(a-y)(a+y)} = \dfrac{a^2 - x^2}{a^2 - y^2}$$ So in the end, we have: $$\dfrac{x^2}{y^2} = \dfrac{a^2 - x^2}{a^2 - y^2}$$ Getting rid of denominators and expanding gives: $$a^2x^2 - x^2y^2 = a^2y^2 - x^2y^2$$ $$a^2x^2 = a^2y^2$$ $$x = y$$ So, this means that $M$ is the midpoint of $XY$, so we are done.